Given:
Mass of crate = 200 kg
Length of crate = 1.25 m
Height of crate = 0.500 m
Angle the stairs make with respect to the floor = 45.0 degrees
Angle the crate is carried = 45 degrees
Let's find the magnitude of the force applied by the person below.
Let's first make a free body diagram representing this situation:
Where:
FL represents the force applied by the person below
Fu represents the force applied by the person above.
[tex]\begin{gathered} \Sigma Fy=0 \\ \\ mg=F_U+F_L \end{gathered}[/tex]
We have:
[tex]mg=200*9.8=1960\text{ N}[/tex]
Now to find the magnitude of force applied by the person below, we have:
[tex]\Sigma m=mg(\frac{1.25}{2}-0.25)cos45-F_U*1.25cos45=0[/tex]
Since the sum of the moments around the bottom must be zero.
Thus, we have:
[tex]\begin{gathered} F_U=\frac{1960(\frac{1.25}{2}-0.25)cos45}{1.25cos45} \\ \\ F_U=\frac{1960(0.375)cos45}{1.25cos45} \\ \\ F_U=588\text{ N} \end{gathered}[/tex]
The force applied by the person above is 588 N.
Thus, the force applied by the person below will be:
[tex]\begin{gathered} F_L=-F_U+mg \\ \\ F_L=-588+200*9.8 \\ \\ F_L=-588+1960 \\ \\ F_U=1372\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of the force applied by the person below is 1372 N.
ANSWER:
1372 N.